1.3 Summation of Series

🔑 Key Concepts and Definitions

Formula 1: Sum of a Constant

$\sum_{r=1}^{n} k = kn$

This represents the sum of a constant $k$ added together $n$ times.

Formula 2: Sum of the First $n$ Natural Numbers

$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$

This represents the sum $1 + 2 + 3 + \ldots + n$.

Formula 3: Sum of the Squares of the First $n$ Natural Numbers

$\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)$

This represents the sum $1^2 + 2^2 + 3^2 + \ldots + n^2$.

Formula 4: Sum of the Cubes of the First $n$ Natural Numbers

$\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2$

This represents the sum $1^3 + 2^3 + 3^3 + \ldots + n^3$.

⚡ Using Standard Results to Find Related Sums

Example 1: Linear Combination

Problem: Find $\sum_{r=1}^{n} (4r + 1)$. Hence determine the value of $\sum_{r=17}^{48} (4r + 1)$.

Solution:

Step 1: Separate the summation

$\sum_{r=1}^{n} (4r + 1) = 4\sum_{r=1}^{n} r + \sum_{r=1}^{n} 1$

Step 2: Apply standard formulae

$= 4 \times \frac{n(n+1)}{2} + n$

Step 3: Simplify

$= 2n(n+1) + n = 2n^2 + 2n + n = 2n^2 + 3n$

Step 4: For the range $r = 17$ to $48$, use subtraction

$\sum_{r=17}^{48} (4r + 1) = \sum_{r=1}^{48} (4r + 1) - \sum_{r=1}^{16} (4r + 1)$

$= (2 \times 48^2 + 3 \times 48) - (2 \times 16^2 + 3 \times 16)$

$= (4608 + 144) - (512 + 48) = 4752 - 560 = 4192$

Example 2: Sum of Odd Numbers (Two Methods)

Problem: Find an expression for $1 + 3 + 5 + 7 + \ldots$ for the first $n$ terms.

Method 1: Direct approach using the $r$-th odd number formula

The $r$-th odd number is $2r - 1$, so:

$\sum_{r=1}^{n} (2r - 1) = 2\sum_{r=1}^{n} r - \sum_{r=1}^{n} 1$

$= 2 \times \frac{n(n+1)}{2} - n = n(n+1) - n = n^2$

Method 2: Using even-odd separation

Consider the sum of the first $2n$ natural numbers, then subtract the even terms:

$\sum_{r=1}^{2n} r - \sum_{r=1}^{n} (2r) = \frac{(2n)(2n+1)}{2} - 2 \times \frac{n(n+1)}{2}$

$= n(2n+1) - n(n+1) = 2n^2 + n - n^2 - n = n^2$

Example 3: Quadratic Expression

Problem: Find an expression in terms of $n$ for $\sum_{r=1}^{n} (3r^2 - 4r + 2)$.

Solution:

$\sum_{r=1}^{n} (3r^2 - 4r + 2) = 3\sum_{r=1}^{n} r^2 - 4\sum_{r=1}^{n} r + \sum_{r=1}^{n} 2$

$= 3 \times \frac{1}{6}n(n+1)(2n+1) - 4 \times \frac{1}{2}n(n+1) + 2n$

$= \frac{1}{2}n(n+1)(2n+1) - 2n(n+1) + 2n$

$= \frac{1}{2}(2n^3 + 3n^2 + n) - 2n^2 - 2n + 2n$

$= n^3 + \frac{3}{2}n^2 + \frac{1}{2}n - 2n^2$

$= n^3 - \frac{1}{2}n^2 + \frac{1}{2}n$

Example 4: Product Expression Requiring Expansion

Problem: Find an expression in terms of $n$ for $\sum_{r=1}^{n} r^2(r - 1)$.

Solution:

Step 1: Expand the general term

$\sum_{r=1}^{n} r^2(r - 1) = \sum_{r=1}^{n} (r^3 - r^2)$

Step 2: Separate and apply formulae

$= \sum_{r=1}^{n} r^3 - \sum_{r=1}^{n} r^2 = \frac{1}{4}n^2(n+1)^2 - \frac{1}{6}n(n+1)(2n+1)$

Step 3: Factor out common terms

$= n(n+1)\left[\frac{n(n+1)}{4} - \frac{2n+1}{6}\right]$

Step 4: Simplify the bracketed expression

$= \frac{n(n+1)}{12}\left[3n(n+1) - 2(2n+1)\right]$

$= \frac{n(n+1)}{12}(3n^2 + 3n - 4n - 2)$

$= \frac{n(n+1)}{12}(3n^2 - n - 2)$

Step 5: Factor the quadratic

$= \frac{n(n+1)}{12}(n - 1)(3n + 2)$

$= \frac{1}{12}n(n+1)(n-1)(3n+2)$